Prove if $p + \sqrt{2} \cdot q = 0$, then $p = q = 0$ Let K be the set 
$ K := \{p + \sqrt{2} . q : p, q \in \mathbb{Q}\} $
and let addition and multiplication be defined on $\mathbb{R}$
Prove if $p + \sqrt{2} \cdot q = 0$, then $p = q = 0$
Hint: use that $ \sqrt{2}$ is irrational
Logically, if $p + \sqrt{2} . q = 0$ this implies $ \sqrt{2} = \frac{-p}{q}$ which is not possible as $ \sqrt{2}$ is irrational. How do I write this as a proof though? 
 A: To me it basically looks like an OK proof (given that you're allowed to actually use that $\sqrt2$ is irrational). What you're using is that $-p$ is rational and dividing two rationals you get a rational number so $-p/q$ is rational.
Perhaps you should be more clear that $p$ and $q$ are rational numbers and especially require that $q\ne 0$ since that's required for $-p/q$ to be defined (on the other hand if $q=0$ the equation would reduce to $p=0$).
I assume that the set $K$ is part of the question and note that $K$ is never used. I think that the question is badly formulated partly for this reason, but also the fact that the requirement for $p$ and $q$ to be rational doesn't follow from an unused set definition (the $p$ and $q$ in the definition of $K$ are bound variables - which means they don't have scope outside the set comprehension).
If you're not allowed to use that $\sqrt2$ is irrational you fix that by just include that as a lemma along with the proof of that.
A: Assuming $pq\neq 0$, suppose $\frac pq$ in it's simplest form can be written as $\frac ab$ i.e. $\gcd(a,b)=1$. Now we have $$\frac ab =-\sqrt 2$$
Squaring both the sides $$ \frac {a^2}{b^2}=2 \implies a^2=2b^2$$
Since RHS is even, LHS must also be even, implying $a=2n$, thus we get $4n^2=2b^2 \implies 2n^2=b^2$.
Now, again using the same argument, we've $b=2m$. But this contradicts our initial assumption that $a$ and $b$ are coprime.
Hence $a=b=0 \implies p=q=0$
A: The formulation of your post seems rather weak and unclear to me.
1.) You define a set $A$ but then there is no reference to this definition. So why do you define it?
2.) When you define $A$ you say "let addition and multiplication be defined on 
$\Bbb R$". What should this mean?  
3.) "Prove if $p + \sqrt{2} \cdot q = 0$, then $p = q = 0$". You tell us nothing about $p$ and $q$. If they are real numbers than your statement is wrong. If they are rational numbers your statement is true.
4.) In the meantime you changed $A$ to $K$ and the definition of $K$ uses variables $p$ and $q$. But this does not change anything, because $p$ and $q$ are bounded variables in this definitions and have no meaning outside of this definition
5.) "Logically...": what means logically here ?
6.) So why is this not possible because $\sqrt 2$ is irrational. More details here! The case $q=0$ must be handled, too.
